NCERT Exemplar Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry and Its Equations

NCERT Exemplar Solutions Class 10 Maths Chapter 8 – Free PDF Download

The NCERT Exemplar Class 10 Maths Chapter 8 Trigonometry and Its Equations is provided here for students to prepare for the board exam. These exemplars problems and solutions are designed by our Maths subject experts with respect to the CBSE Syllabus (2023-2024). Students can make use of the exemplar to strengthen their exam preparation. Moreover, they will find answers to all the tough questions, and will be able to prepare efficiently, as well as deliver better performance in the exam.

NCERT Exemplar Class 10 Maths Chapter 8 focuses mainly on helping students with learning topics like trigonometric identities and trigonometric ratios. In this chapter, they will learn to solve problems based on trigonometric ratios for specific and complementary angles and establish identities for the trigonometric ratios. However, students have to be thorough with the concepts mentioned in the chapter to solve problems without any difficulties. Click here to solve exemplars for all chapters of Maths Standard 10. This chapter covers the following topics of trigonometry:

• Finding the trigonometric ratios of the angle
• Finding the trigonometric ratios of some specific angles, such as 30°, 45°, 60°, 90° and 0°
• Trigonometric ratios of complementary angles, like sin(90° – A)
• Proofs based on trigonometric identities

Download the PDF of NCERT Exemplar Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry and Its Equations

Access Answers to the NCERT Exemplar Class 10 Maths Chapter 8

Exercise 8.1 Page No: 89

Choose the correct answer from the given four options:

1. If cos A = 4/5, then the value of tan A is

(A) 3/5 (B) ¾ (C) 4/3 (D) 5/3

Solution:

(B) 3/4

According to the question,

cos A = 4/5 …(1)

We know,

tan A = sinA/cosA

To find the value of sin A,

We have the equation,

sin² θ +cos² θ =1

So, sin θ = √ (1-cos² θ)

Then,

sin A = √ (1-cos² A) …(2)

sin² A = 1-cos² A

sin A = √(1-cos² A)

Substituting equation (1) in (2),

We get,

Sin A = √(1-(4/5)²)

= √(1-(16/25))

= √(9/25)

= ¾

Therefore,

2. If sin A = ½ , then the value of cot A is

(A) √3 (B) 1/√3 (C) √3/2 (D) 1

Solution:

(A) √3

According to the question,

Sin A = ½ … (1)

We know that,

 … (2)

To find the value of cos A.

We have the equation,

sin² θ +cos² θ =1

So, cos θ = √(1-sin² θ)

Then,

cos A = √(1-sin² A) … (3)

cos² A = 1-sin² A

cos A = √ (1-sin² A)

Substituting equation 1 in 3, we get,

cos A = √(1-1/4) = √(3/4) = √3/2

Substituting values of sin A and cos A in equation 2, we get

cot A = (√3/2) × 2 = √3

3. The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is

(A) – 1 (B) 0 (C) 1 (D) 3 2

Solution:

(B) 0

According to the question,

We have to find the value of the equation,

cosec(75°+θ) – sec(15°-θ) – tan(55°+θ) + cot(35°-θ)

= cosec[90°-(15°-θ)] – sec(15°-θ) – tan(55°+θ) + cot[90°-(55°+θ)]

Since, cosec (90°- θ) = sec θ

And, cot(90°-θ) = tan θ

We get,

= sec(15°-θ) – sec(15°-θ) – tan(55°+θ) + tan(55°+θ)

= 0

4. Given that sinθ = a b , then cosθ is equal to

(A) b/√(b²– a²) (B) b/a (C) √(b²-a²)/b (D) a/√(b²-a²)

Solution:

(C) √(b² – a²)/b

According to the question,

sin θ =a/b

We know, sin² θ +cos² θ =1

sin² A = 1-cos² A

sin A = √(1-cos² A

So, cos θ = √(1-a²/b² ) = √((b²-a²)/b² ) = √(b²-a² )/b

Hence, cos θ = √(b² – a² )/b

5. If cos (α + β) = 0, then sin (α – β) can be reduced to

(A) cos β (B) cos 2β (C) sin α (D) sin 2α

Solution:

(B) cos 2β

According to the question,

cos(α+β) = 0

Since, cos 90° = 0

We can write,

cos(α+β)= cos 90°

By comparing cosine equation on L.H.S and R.H.S,

We get,

(α+β)= 90°

α = 90°-β

Now we need to reduce sin (α -β ),

So, we take,

sin(α-β) = sin(90°-β-β) = sin(90°-2β)

sin(90°-θ) = cos θ

So, sin(90°-2β) = cos 2β

Therefore, sin(α-β) = cos 2β

6. The value of (tan1° tan2° tan3° … tan89°) is

(A) 0 (B) 1 (C) 2 (D) ½

Solution:

(B) 1

tan 1°. tan 2°.tan 3° …… tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.tan 45°.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

Since, tan 45° = 1,

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan(90°-44°).tan(90°-43°)…tan(90°-3°). tan(90°-2°).tan(90°-1°)

Since, tan(90°-θ) = cot θ,

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.cot 44°.cot 43°…cot 3°.cot 2°.cot 1°

Since, tan θ = (1/cot θ)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1. (1/tan 44^o). (1/tan 43^o)… (1/tan 3^o). (1/tan 2^o). (1/tan 1^o)

= 1

Hence, tan 1°.tan 2°.tan 3° …… tan 89° = 1

7. If cos 9α = sinα and 9α < 90°, then the value of tan5α is

(A) 1/√3 (B) √3 (C) 1 (D) 0

Solution:

(C) 1

According to the question,

cos 9∝ = sin ∝ and 9∝<90°

i.e. 9α is an acute angle

We know that,

sin(90°-θ) = cos θ

So,

cos 9∝ = sin (90°-∝)

Since, cos 9∝ = sin(90°-9∝) and sin(90°-∝) = sin∝

Thus, sin (90°-9∝) = sin∝

90°-9∝ =∝

10∝ = 90°

∝ = 9°

Substituting ∝ = 9° in tan 5∝, we get,

tan 5∝ = tan (5×9) = tan 45° = 1

∴, tan 5∝ = 1

Exercise 8.2 Page No: 93

Write ‘True’ or ‘False’ and justify your answer in each of the following:

1. tan 47^o/cot 43 ° = 1

Solution:

True

Justification:

Since, tan (90° -θ) = cot θ

2. The value of the expression (cos²23° – sin²67°) is positive.

Solution:

False

Justification:

Since, (a²-b²) = (a+b)(a-b)

cos² 23° – sin² 67° =(cos 23°+sin 67°)(cos 23°-sin 67°)

= [cos 23°+sin(90°-23°)] [cos 23°-sin(90°-23°)]

= (cos 23°+cos 23°)(cos 23°-cos 23°) (∵sin(90°-θ) = cos θ)

= (cos 23°+cos 23°).0

= 0, which is neither positive nor negative

3. The value of the expression (sin 80° – cos 80°) is negative.

Solution:

False

Justification:

We know that,

sin θ increases when 0° ≤ θ ≤ 90°

cos θ decreases when 0° ≤ θ ≤ 90°

And (sin 80°-cos 80°) = (increasing value-decreasing value)

= a positive value.

Therefore, (sin 80°-cos 80°) > 0.

4. √((1– cos²θ) sec² θ)= tan θ

Solution:

True

Justification:

5. If cosA + cos²A = 1, then sin²A + sin⁴A = 1.

Solution:

True

Justification:

According to the question,

cos A+cos² A = 1

i.e., cos A = 1- cos² A

Since,

sin² θ+cos² θ = 1

sin² θ = 1- cos² θ)

We get,

cos A = sin² A …(1)

Squaring L.H.S and R.H.S,

cos² A = sin⁴ A …(2)

To find sin²A+sin⁴ A=1

Adding equations (1) and (2),

We get

sin²A + sin⁴ A= cos A + cos² A

Therefore, sin²A+ sin⁴ A = 1

6. (tan θ + 2) (2 tan θ + 1) = 5 tan θ + sec² θ.

Solution:

False

Justification:

L.H.S = (tan θ+2) (2 tan θ+1)

= 2 tan² θ + tan θ + 4 tan θ + 2

= 2 tan²θ+5 tan θ+2

Since, sec² θ – tan² θ = 1, we get, tan²θ = sec² θ-1

= 2(sec² θ-1) +5 tan θ+2

= 2 sec² θ-2+5 tan θ+2

= 5 tan θ+ 2 sec² θ ≠R.H.S

∴, L.H.S ≠ R.H.S

Exercise 8.3 Page No: 95

Prove the following (from Q.1 to Q.7):

1. sin θ/(1+cos θ) + (1+ cos θ)/sin θ = 2cosec θ

Solution:

L.H.S=

R.H.S

Hence proved.

2. tan A/(1+secA) – tan A/(1-secA) = 2cosec A

Solution:

L.H.S:

Since,

sec²A – tan²A = 1

sec²A – 1 = tan²A

= R.H.S

Hence proved.

3. If tan A = ¾, then sinA cosA = 12/25

Solution:

According to the question,

tan A = ¾

We know,

tan A = perpendicular/ base

So,

tan A = 3k/4k

Where,

Perpendicular = 3k

Base = 4k

Using Pythagoras Theorem,

(hypotenuse)² = (perpendicular)² + (base)²

(hypotenuse)² = (3k)²+ (4k)² = 9k²+16k² = 25k²

hypotenuse = 5k

To find sin A and cos A,

Hence, proved.

4. (sin α + cos α) (tan α + cot α) = sec α + cosec α

Solution:

L.H.S:

(sin α + cos α) (tan α + cot α)

As we know,

= R.H.S

Hence, proved.

5. (√3+1) (3 – cot 30°) = tan³ 60° – 2 sin 60°

Solution:

L.H.S: (√3 + 1) (3 – cot 30°)

= (√3 + 1) (3 – √3) [∵cos 30° = √3]

= (√3 + 1) √3 (√3 – 1) [∵(3 – √3) = √3 (√3 – 1)]

= ((√3)²– 1) √3 [∵ (√3+1)(√3-1) = ((√3)² – 1)]

= (3-1) √3

= 2√3

Similarly solving R.H.S: tan³ 60° – 2 sin 60°

Since, tan 60^o = √3 and sin 60^o = √3/2,

We get,

(√3)³ – 2.(√3/2) = 3√3 – √3

= 2√3

Therefore, L.H.S = R.H.S

Hence, proved.

6. 1 + (cot² α/1+cosec α) = cosec α

Solution:

And, we know that,

7. tan θ + tan (90° – θ) = sec θ sec (90° – θ)

Solution:

L.H.S=

Since, tan (90° – θ) = cot θ

tan θ + tan (90° – θ)= tan θ + cot θ

Exercise 8.4 Page No: 99

1. If cosecθ + cotθ = p, then prove that cosθ = (p² – 1)/ (p² + 1).

Solution:

According to the question,

cosec θ + cot θ = p

Since,

Hence, proved.

2. Prove that √(sec² θ + cosec² θ) = tan θ + cot θ

Solution:

L.H.S=

√ (sec² θ + cosec² θ)

Since,

= R.H.S

Hence, proved.

3. The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.

Solution:

Let PR = h meter, be the height of the tower.

The observer is standing at point Q such that, the distance between the observer and tower is QR = (20+x) m, where

QR = QS + SR = 20 + x

∠PQR = 30°

∠ PSR = θ

In ∆PQR,

Rearranging the terms,

We get 20 +x = √3h

⇒ x = √3h – 20 …eq.1

In ∆PSR,

tan θ = h/x

Since, angle of elevation increases by 15^o when the observer moves 20 m towards the tower.

We have,

θ = 30° + 15° = 45°

So,

tan 45^o = h/x

⇒ 1 = h/x

⇒ h = x

Substituting x=h in eq. 1, we get

h = √3 h – 20

⇒ √3 h – h = 20

⇒ h (√3 – 1) = 20

= 10 (√3 + 1)

Hence, the required height of the tower is 10 (√3 + 1) meter.

4. If 1 + sin²θ = 3sinθ cosθ , then prove that tanθ = 1 or ½.

Solution:

Given: 1+sin² θ = 3 sin θ cos θ

Dividing L.H.S and R.H.S equations with sin² θ,

We get,

cosec² θ + 1 = 3 cot θ

Since,

cosec² θ – cot² θ = 1 ⇒ cosec² θ = cot² θ +1

⇒ cot² θ +1+1 = 3 cot θ

⇒ cot² θ +2 = 3 cot θ

⇒ cot² θ –3 cot θ +2 = 0

Splitting the middle term and then solving the equation,

⇒ cot² θ – cot θ –2 cot θ +2 = 0

⇒ cot θ(cot θ -1)–2(cot θ +1) = 0

⇒ (cot θ – 1)(cot θ – 2) = 0

⇒ cot θ = 1, 2

Since,

tan θ = 1/cot θ

tan θ = 1, ½

Hence, proved.

5. Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.

Solution:

Given: sin θ +2 cos θ = 1

Squaring on both sides,

(sin θ +2 cos θ)² = 1

⇒ sin² θ + 4 cos² θ + 4sin θcos θ = 1

Since, sin² θ = 1 – cos² θ and cos² θ = 1 – sin² θ

⇒ (1 – cos² θ) + 4(1 – sin² θ) + 4sin θcos θ = 1

⇒ 1 – cos² θ + 4 – 4 sin² θ + 4sin θcos θ = 1

⇒ – 4 sin² θ – cos² θ + 4sin θcos θ = – 4

⇒ 4 sin² θ + cos² θ – 4sin θcos θ = 4

We know that,

a²+ b² – 2ab = ( a – b)²

So, we get,

(2sin θ – cos θ)² = 4

⇒ 2sin θ – cos θ = 2

Hence proved.

6. The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √st .

Solution:

Let BC = s; PC = t

Let height of the tower be AB = h.

∠ABC = θ and ∠APC = 90° – θ

(∵ the angle of elevation of the top of the tower from two points P and B are complementary)

⇒ h² = st

⇒ h = √st

Hence the height of the tower is √st.

7. The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower.

Solution:

Let SQ = h be the tower.

∠SPQ = 30° and ∠SRQ = 60°

According to the question, the length of shadow is 50 m long hen angle of elevation of the sun is 30° than when it was 60°. So,

PR = 50 m and RQ = x m

So in ∆SRQ, we have

⇒ 50√3+h = 3h

⇒ 50√3 = 3h – h

⇒ 3h – h = 50√3

⇒ 2h = 50√3

⇒ h = (50√3)/2

⇒ h = 25√3

Hence, the required height is 25√3 m.

8. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are α and β, respectively. Prove that the height of the tower is [h tan α/(tan β – tan α)].

Solution:

Given that a vertical flag staff of height h is surmounted on a vertical tower of height H(say),

such that FP = h and FO = H.

The angle of elevation of the bottom and top of the flag staff on the plane is ∠PRO = α and ∠FRO = β respectively

In ∆PRO, we have

Hence, proved.

9. If tanθ + secθ = l, then prove that secθ = (l² + 1)/2l.

Solution:

Given: tan θ+ sec θ = l …eq. 1

Multiplying and dividing by (sec θ – tan θ) on numerator and denominator of L.H.S,

So, sec θ – tan θ = 1 …eq.2

Adding eq. 1and eq. 2, we get

(tan θ + sec θ) + (sec θ – tan θ) = 1

Hence, proved.

Class 10 students are advised to solve sample papers and previous years’ question papers to have an idea of the types of questions asked from the chapter Trigonometry. Also, find our online reading materials, such as notes, exemplar books, 10th Class Maths NCERT Solutions and question papers in downloadable PDF format to have a quick revision before the board exam.

Download BYJU’S – The Learning App to get personalised video lessons explaining Maths topics. Experience a new method of learning at BYJU’S to understand concepts effortlessly.

Also, Read